Question

If \[ f(x) = \begin{cases} \dfrac{\sin(\cos x) - \cos x}{(\pi - 2x)^3}, & x \ne \dfrac{\pi}{2} \\ k, & x = \dfrac{\pi}{2} \end{cases} \] is continuous at \( x = \dfrac{\pi}{2} \), then \( k \) is equal to:

• A. 0
• B. \(-\frac{1}{6}\)
• C. \(-\frac{1}{24}\)
• D. \(-\frac{1}{48}\)

Hint:

Use series expansion or L'Hôpital's rule for limits involving trigonometric functions.

Answer 1

The Correct Option is D

Step 1: Concept:
For continuity at \( x = \frac{\pi}{2} \), we must have: \[ k = \lim_{x \to \frac{\pi}{2}} f(x) \]
Step 2: Detailed Explanation:
Let, \[ t = \frac{\pi}{2} - x \Rightarrow t \to 0 \] Then, \[ \cos x = \cos\left(\frac{\pi}{2} - t\right) = \sin t \]
Numerator:
\[ \sin(\cos x) - \cos x = \sin(\sin t) - \sin t \] Using expansion: \[ \sin(\sin t) = \sin t - \frac{(\sin t)^3}{6} + \cdots \] So, \[ \sin(\sin t) - \sin t \approx -\frac{(\sin t)^3}{6} \] Since \( \sin t \approx t \), \[ \sin(\sin t) - \sin t \approx -\frac{t^3}{6} \]
Denominator:
\[ (\pi - 2x)^3 = (2t)^3 = 8t^3 \]
Now,
\[ f(x) \approx \frac{-t^3/6}{8t^3} = -\frac{1}{48} \]
Thus, \[ \lim_{x \to \frac{\pi}{2}} f(x) = -\frac{1}{48} \]
Step 3: Final Answer:
\[ \boxed{k = -\frac{1}{48}} \]